Rapid Recall
- Which problems are you ready to present?
- Which problems did you sincerely attempt
- Given the vectors $\vec F$ and $\vec d$ on the board, draw the projection of $\vec F$ onto $\vec d$.
- If I know that $\vec F=(0,-3)$ and $\vec F_{\parallel \vec d} = (1,-2)$, then state $\vec F_{\perp \vec d}$.
- The force $\vec F = (3,5)$ acts on an object which undergoes a displacement $(-2,1)$. Find the work done by $\vec F$ though this displacement.
Group problems
- The projection of $\vec F$ onto $\vec d$ is $(3,0)$. Give two vectors $\vec F$ and $\vec d$ so that this can occur.
- Give a different pair $\vec F$ and $\vec d$ with the same projection.
- Repeat the previous problem, assuming that $\text{proj}_{\vec d}{\vec F} = (-1,2)$.
- Let $x=2u+3$ and $y=4v-5$. Complete the $u,v,x,y$ table below, and then construct a graph of both $u^2+v^2=1$ (in the $uv$-plane) and the corresponding equation in the $xy$-plane. $$ \begin{array}{c|c|c|c} u&v&x&y\\\hline 0&0&3&-5\\ 1&0&5&-5\\ 0&1&&\\ -1&0&&\\ 0&-1&&\\ \end{array} $$
- Draw the three circles $x^2+y^2=1$, $(x-2)^2+(y-5)^2=1$, and $(x+2)^2+(y+3)^2=1$.
- Draw $\left(\frac{x}{3}\right)^2+\left(\frac{y}{5}\right)^2=1$.
- Draw $\ds \frac{(x-2)^2}{16}+\frac{(y-3)^2}{9}=1$.
- Draw $\ds \frac{(x-2)^2}{16}-\frac{(y-3)^2}{9}=1$.
- Draw $\ds \frac{(y-3)^2}{9}-\frac{(x-2)^2}{16}=1$.
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